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Irreversibility Physics 313 Professor Lee Carkner Lecture 16

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Exercise #15 Carnot Engine Power of engine = 1 – Q H /Q L = W/Q H Source temp = 1 – T L /T H Max refrigerator COP For a Carnot refrigerator operating between the same temperatures: Since K < K C (8.2<9.9), refrigerator is possible

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Entropy Entropy (S) defined by heat and temperature Total entropy around a closed reversible path is zero Can write heat in terms of entropy: dQ = T dS

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General Irreversibility Since S = S f - S i S f > S i This is true only for the sum of all entropies Since only irreversible processes are possible, Entropy always increases

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Reversible Processes Consider a heat exchange between a system and reservoir at temperature T So: dS s = +dQ/T dS r = - dQ/T For a reversible process the total entropy change of the universe is zero

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Irreversible Processes How do you compute the entropy change for an irreversible process? What is the change in entropy for specific irreversible processes?

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Isothermal W to U Friction or stirring of a system in contact with a heat reservoir The only change of entropy is heat Q (=W) absorbed by the reservoir S = W/T

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Adiabatic W to U Friction or stirring of insulated substance System will increase in temperature S = dQ/T = C P dT/T = C P ln (T f /T i )

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Heat Transfer Transferring heat from high to low T reservoir For any heat reservoir S = Q/T S for cool reservoir = + Q/T C Assumes no other changes in any other system

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Free Expansion Gas released into a vacuum Replace with a reversible isothermal expansion Thus, (dQ/T) = (nRdV/V) Note: Entropy increases even though temperature does not change

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Entropy Change of Solids Solids (and most liquids) are incompressible We can thus write dQ as CdT and dS as (C/T)dT If we approximate C as being constant with T Note: If C is not constant with T, need to know (and be able to integrate) C(T)

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General Entropy Changes For fluids that under go a change in T, P or V we can find the entropy change of the system by finding dQ For example ideal gas: dQ = C P dT – VdP dQ = C V dT + PdV These hold true for any continuous process involving an ideal gas with constant C

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Notes on Entropy Processes can only occur such that S increases Entropy is not conserved The degree of entropy increase indicates the degree of departure from the reversible state

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Use of Entropy How can the second law be used? Example: total entropy for a refrigerator S (reservoir) = (Q + W) /T H The sum of all the entropy changes must be greater than zero:

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Use of Entropy (cont.) We can now find an expression for the work: Thus the smallest value for the work is: Thus for any substance we can look up S 1 -S 2 for a given Q and find out the minimum amount of work needed to cool it

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